Question

If the shear force diagram for a beam is a triangle with the length of the beam as its base, the beam is

• A. A cantilever beam with a concentrated load at its free end
• B. A cantilever beam with uniformly distributed load over its whole span
• C. A simply supported beam with a concentrated load at its mid point
• D. A simply supported beam with uniformly distributed load over its whole span

Hint:

A triangular shear force diagram typically indicates a linearly varying load, such as a uniformly distributed load on a cantilever beam.

Answer 1

The Correct Option is B

Step 1: Understand the shear force diagram.
The shear force diagram (SFD) is a triangle with the length of the beam as its base. This means the shear force varies linearly from a maximum value at one end to zero at the other, forming a triangular shape. Step 2: Analyze the shear force for each option.
(1) A cantilever beam with a concentrated load at its free end:
For a cantilever beam (fixed at one end, free at the other) with a concentrated load \( P \) at the free end:
At the free end, shear force \( V = P \).
At the fixed end, shear force \( V = P \).
The shear force is constant along the length, so the SFD is a rectangle, not a triangle. Incorrect.
(2) A cantilever beam with uniformly distributed load over its whole span:
For a cantilever beam with a uniformly distributed load (UDL) \( w \, \text{N/m} \) over length \( L \):
At the free end (\( x = 0 \)), shear force \( V = 0 \).
At the fixed end (\( x = L \)), shear force \( V = wL \) (total load).
Shear force varies linearly: \( V(x) = w x \), from 0 at the free end to \( wL \) at the fixed end.
The SFD is a triangle with base \( L \), starting at 0 and increasing to \( wL \). Matches the description.
(3) A simply supported beam with a concentrated load at its mid point:
For a simply supported beam with a concentrated load \( P \) at the midpoint:
Reactions at supports: \( R_A = R_B = \frac{P}{2} \).
From \( x = 0 \) to \( x = \frac{L}{2} \), shear force \( V = \frac{P}{2} \).
At \( x = \frac{L}{2} \), shear force drops by \( P \), so \( V = \frac{P}{2} - P = -\frac{P}{2} \).
From \( x = \frac{L}{2} \) to \( x = L \), shear force \( V = -\frac{P}{2} \).
The SFD is a step function (rectangle, not a triangle). Incorrect.
(4) A simply supported beam with uniformly distributed load over its whole span:
For a simply supported beam with UDL \( w \, \text{N/m} \):
Reactions: \( R_A = R_B = \frac{wL}{2} \).
At \( x = 0 \), shear force \( V = \frac{wL}{2} \).
At \( x = \frac{L}{2} \), shear force \( V = \frac{wL}{2} - w \cdot \frac{L}{2} = 0 \).
At \( x = L \), shear force \( V = \frac{wL}{2} - wL = -\frac{wL}{2} \).
The SFD is a straight line from \( \frac{wL}{2} \) to \( -\frac{wL}{2} \), but not a triangle with base as the beam length (it crosses zero at the midpoint). Incorrect. Step 3: Select the correct answer.
The shear force diagram being a triangle with the beam length as its base matches a cantilever beam with a uniformly distributed load, option (2).